Question

Let $S$ and $T$ be two arrays of $m$ and $n$ elements, respectively. Write an algorithm that finds all the common elements and stores them in an array $U$ Show that this can be done in $\mathrm{\Theta }(n+m)$ time.

Short answer

The algorithm compares elements from the two arrays using two pointers. If a common element is found, it is stored in array U. The time complexity is $\mathrm{\Theta }(n+m)$ because in the worst case, every element in each array will be compared once, leading to $n+m$ comparisons.

Step-by-step solution

Initialize arrays and variables

Start by initializing the two input arrays $S$ and $T$, as well as the output array $U$. Also initialize two pointers, $i$ and $j$, as 0. The pointers will start navigating from the beginning of each array.

Describing the algorithm

While both pointers $i$ and $j$ are within their respective array lengths, compare $S[i]$ and $T[j]$. If $S[i]$ is equal to $T[j]$, meaning a common element is found, add this element to $U$ and increment both pointers. If $S[i]$ is less than $T[j]$, increment $i$, else increment $j$.

Understand the Time Complexity

The time complexity of this algorithm is $\mathrm{\Theta }(n+m)$ because in the worst-case scenario, every element in each array has to be compared once. This requires $n+m$ comparisons. So the algorithm's time complexity is linear with respect to the sum of the lengths of the two arrays.

Key concepts

Array Comparison

When working with data structures, particularly arrays, the operation of finding common elements between them is known as array comparison. In layman's terms, this involves going through each element of one array and checking if it exists in the other array. The result is often a new array containing only those elements present in both initial arrays.

For the given exercise, the arrays to be compared are called $S$ and $T$, each containing $m$ and $n$ elements respectively. The task is to produce a new array $U$ that will hold the elements common to both. It's essential to compare each item from the first array with the other effectively without unnecessary repetitions.

One way to do this is by using two pointers, starting at the beginning of each array. These pointers move through the arrays and are adjusted based on comparisons of their respective elements. This technique ensures that each element is looked at and compared efficiently—allowing for the process of merging common elements into the output array to be performed seamlessly.

Time Complexity

Time complexity is a way to express how the running time of an algorithm increases with the size of the input. It's crucial for understanding how efficient an algorithm is, especially when dealing with large datasets. The notation $\mathrm{\Theta }$ is used to denote that the running time grows proportionally to the size of the input.

In the case of our array comparison algorithm, the exercise states and asks you to confirm that it can be done in $\mathrm{\Theta }(n+m)$ time. This signifies that the time needed to find all common elements scales linearly with the combined size of the two input arrays. Since each array is navigated independently and at most once, the algorithm efficiently minimizes operations and avoids unnecessary repetition.

If you visualize the comparison as a single walk through the entire lengths of both arrays, you can see that no element is ever revisited, which maintains the lean runtime of the process. This linear relationship is beneficial for working with large arrays because it implies predictability in performance, allowing programmers and engineers to estimate the resources needed for the task.

Pseudocode

Pseudocode is a simplified, half-coded version of an algorithm written in plain language. It doesn’t adhere to the strict syntax of any particular programming language, but outlines the logic and steps necessary to perform a task or solve a problem. Because of its simplicity, pseudocode is easily understandable and serves as a bridge between coming up with a concept and coding it.

Let's illustrate this with the array comparison problem:

Begin Algorithm

Initialize two pointers, $i$ and $j$, to 0.
While $i<m$ and $j<n$:
If $S[i]==T[j]$:

• Add $S[i]$ to array $U$
• Increment both $i$ and $j$

If $S[i]<T[j]$:

• Increment $i$

Else:

• Increment $j$

End Algorithm

This pseudocode distills the algorithm into clear steps devoid of complicated code, which is essential for understanding the methodology of the solution before diving into programming it. This step-by-step approach guides developers in how to build the final code, allowing for optimizations and error-checking along the way.